A water pipe tapers down from an initial radius of R 1 = 0.2 m to a final radius

Question

A water pipe tapers down from an initial radius of R1 = 0.2 m to a final radius of R2 = 0.1 m. The water flows at a velocity v1 = 0.82 m/s in the larger section of pipe.

1)

What is the volume flow rate of the water?

m3/s  

2)

What is the velocity of the water in the smaller section?

m/s  

3)

Using this water supply, how long would it take to fill up a swimming pool with a volume of V = 158 m3? (give your answer in minutes)

min  

4)

The water pressure in the center of the larger section of the pipe is P1 = 243120 Pa. Assume the density of water is 103 kg/m3.

What is the pressure in the center of the smaller section of the pipe?

Explanation / Answer

1) volume flo rate = A*v = 3.14*0.2^2*0.82 = 0.103 m^3/s

2) A1v1 = A2v2

v2 = 0.2^2*0.82/0.1^2 = 3.28 m/s

3) t = volume/volume flow rate = 158/0.103 = 25.56 min

4) p2 = p1+1/2rho(v1^2-v2^2) = 237760 Pa

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